In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = $72^{\circ}$ , find $∠$ PTR.

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Solution

$∠$ OQR = $∠$ ORQ (OR=OQ (radius))

But $∠$ ROP = $∠$ OQR + $∠$ ORQ (Exterior angle theorem)

$72^{\circ}$ = 2 $∠$ OQR

$∠$ OQR = $36^{\circ}$

In $△$ QPT

$∠$ PTQ = $180^{\circ}$ - $∠$ PQT - $∠$ QPT

$∠$ PTQ = $180^{\circ}$ – $36^{\circ}$ - $90^{\circ}$

$∠$ PTQ = $54^{\circ}$

Therefore, $∠$ PTR = $54^{\circ}$

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In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = 72∘, find ∠PTR.

∠OQR = ∠ORQ (OR=OQ (radius)) But ∠ROP = ∠OQR + ∠ORQ (Exterior angle theorem) 72∘ = 2 ∠OQR ∠OQR = 36∘ In △QPT ∠PTQ = 180∘ - ∠PQT - ∠QPT ∠PTQ = 180∘ – 36∘ - 90∘ ∠PTQ = 54∘ Therefore, ∠PTR = 54∘

In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = $72^{\circ}$ , find $∠$ PTR.